by Alex Bellos
The puzzle is tantalizing. The two questions are almost identical, but have drastically different answers, one of which seems to go against common sense. If we know that Mr Smith has a son, but we don’t know whether this son is the older or the younger child, the chance of him having two sons is 1/3. But if we are told that the older (or indeed the younger) child is a son, the chance of him having two sons rises to 1/2. It seems paradoxical that specifying the birth order makes a difference, because we know with 100 per cent certainty that the boy is either the older or the younger.
Scientific American had barely hit the shops when the complaints started to come in. Readers pointed out that the Mr Smith question, as posed by Gardner, was ambiguous, since the given answer depended on how the information about his children was obtained. Gardner admitted the error and wrote a follow-up column about the perils of ambiguity in probability problems.
To make the Mr Smith question watertight we need to state that Mr Smith was chosen at random from the population of all families with exactly two children. If this is the case – and it is arguably how most people understand the question – then the probability that he has two boys is indeed 1/3. But imagine, on the other hand, if Mr Smith was chosen at random from the population of all two-children families and told to say ‘at least one is a boy’ if he has two boys, ‘at least one is a girl’ if he has two girls and if he has one of each to choose at random whether to say he has at least one boy or at least one girl. In this case, the probability he has two boys is 1/2. (The chance of having two boys is 1/4, the chance of two girls is 1/4 and the chance of one of each is 1/2. So the chance he has two boys is 1/4 + [1/2 x 1/2] = 1/2.)
The pitfalls of ambiguity mean that problems like this one must clearly state the process by which the available information is obtained. A further issue is plausibility. What are the real-life scenarios in which a parent of two children would meaningfully say ‘at least one of my children is a boy’ without specifying in any way which child is the boy? These next questions explore this issue.
Tread carefully, boys and girls.
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PROBLEMS WITH SIBLINGS
For these questions assume that the parents are chosen at random from the population of families with exactly two children.
[1] Albert has two children. He is asked to complete the following form.
Of the following three choices, circle exactly one that is true for you:
My older child is a boy
My younger child is a boy
Neither child is a boy
If both are boys, flip a coin to decide which of the first two choices to circle.
Albert circles the first sentence (‘My older child is a boy’). What is the probability that both of his children are boys?
[2] A reporter sees Albert’s completed form, and writes in the newspaper:
Albert has exactly two children, the oldest of which is a son. Is a reader of the newspaper correct to deduce that the chance of Albert having two sons is 1 in 2?
[3] Beth has two children.
You: Can you think of one of your children?
Beth: OK, I have one in mind.
You: Is that child a girl?
Beth: Yes.
What do you estimate are the chances that Beth has two girls?
[4] All you know about Caleb is that he has two children, at least one of which is a girl. We can imagine asking him whether his older child is a girl, or whether his younger child is a girl, and we know his answer to at least one of these questions will be ‘Yes’. Does this insight alter the chances of him having two girls from 1 in 3 to 1 in 2?
[5] A year ago, I learned that Caleb has exactly two children. I asked him one of the following two questions: ‘Is your older child a girl?’ or ‘Is your younger child a girl?’ I know that his answer was ‘Yes’, but I can’t remember which question I asked! It’s a 50/50 chance I asked either question. What are the chances he has two girls?
These problems were set by the brothers Tom and Michael Starbird.
(Question to Mrs Starbird: You have two sons. At least one is a boy. What is the probability that both will become mathematicians?)
Tom is a veteran NASA scientist who has worked on many space missions and continues to help operate the Mars Curiosity rover, while Michael is a professor of mathematics at the University of Texas.
Michael also presents home-learning lecture courses. In the early 2000s, when he was preparing one on probability, the brothers came up with a twist on Gardner’s two-boy problem in which Mr Smith has two children, one of whom was a boy born on a Tuesday. Remarkably, once you mention the day of the week, the probability of Mr Smith having two boys is no longer 1⁄2 or 1⁄3 but somewhere in between. (Assuming that Mr Smith is randomly chosen from the population of all two-children families). The answer seems incomprehensible. How on Earth can knowing the day of the week make a difference to the probability that both children are boys, since the boy under discussion is equally likely to have been born on a Tuesday than he is likely to have been born on any other day?
The Starbirds did not realise quite how their Tuesday-boy problem would cause a trail of misery and irritation. ‘Many people resist the idea that apparent irrelevancies can have an impact on probabilities,’ says Michael. ‘People are so upset by it. Of course, surprises and counter-intuitive results are good, so I like people to be startled, but in this case the “aha” moment sometimes never comes.’
Tom says that two of his colleagues at the Jet Propulsion Laboratory reacted to the puzzle with ‘close to genuine anger. There is something about the puzzle that is seriously disconcerting to some people.’ He wonders if people get emotionally involved in this type of puzzle because it challenges their basic mathematical confidence – ‘there may be a subconscious aspect that is upsetting, along the lines of: if my intuition is so wrong for such an easy-to-state problem, am I making errors all the time in other areas?’
The solution to (and further discussion of) the Tuesday-boy problem is provided in the answer to the next question.
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THE GIRL BORN IN AN EVEN YEAR
Doris has two children, at least one of whom is a girl born in an even-numbered year. What is the probability that both of her children are girls?
Assume that Doris is chosen at random from the population of two-child families and that babies are equally likely to be born in odd- or even-numbered years.
Here’s another puzzle about siblings that tricks our intuition. This time – you may be pleased to hear – their genders and birthdays are irrelevant.
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THE TWYNNE TWINS
A class of 30 students – including the Twynne twins – lines up in single file in the lunch queue. The students select their positions randomly, that is, each student has an equal chance of being in any position. The line is single-file, so one of the Twynne twins will be ahead of the other in the queue. Call the Twynne twin nearer the front of the queue the ‘first’ twin in the queue.
If you had to bet that the first Twynne twin will be in a particular position in the line (as in, are they first, second, third…) which position would you pick as most likely? In other words, if the 30 students line up in a random order every lunchtime, where in the line would the first twin be most often?
Since the students select their positions randomly, in the long run each student will appear in every position roughly the same number of times. What’s counter-intuitive, however, is that the Twynne twin nearer the front of the queue is more likely to appear in one position than in any other.
The development of probability as a mathematical field led, in the nineteenth century, to the birth of modern statistics, the collection, analysis and interpretation of data. One important idea from statistics is the ‘average’ in a set of data. You may remember from school that the most common types of average are the mean (add up all the values and divide them by the number of terms), the median (the middle value if they are all listed in numerical
order) and the mode (the most common value). Another familiar term when looking at data is the range, which is the difference between the highest and lowest values.
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A JAB OF MMMR
Find two sets of five numbers for which the mean, median, mode and range are all 5. Use only whole numbers.
Averages often mislead. It all depends how you aggregate the data.
118
LIES AND STATISTICS
A headmaster is in charge of a junior school and an upper school. Across all pupils the median pupil, ranked by grade, gets a C in maths.
The headmaster introduces a new maths syllabus. The following year the median pupil’s score has reduced to grade D.
Can you devise a scenario in which the headmaster has in fact improved everyone’s grades?
The headmaster has all his pupils’ scores. Often, however, statisticians do not have all the data and must make estimates based on probabilities.
119
THE LONELINESS OF THE LONG-DISTANCE RUNNER
You know that there is a race on in your neighbourhood. You look out of your window and a runner passes by with the number 251 on their chest.
Given that the race organisers number the runners consecutively from 1, and that this runner is the only one you have seen, what is your best guess for how many runners are in the race? Take ‘best guess’ to mean the estimate that makes your observation the most likely.
Continuing our sporting interlude …
120
THE FIGHT CLUB
In order to become accepted into the prestigious Golden Cage Club, you need to win two cage fights in a row. You will fight three bouts in total. Your adversaries will be the fearsome Beast and the considerably less fearsome Mouse. In fact, you estimate that you only have a 2⁄5 chance of beating Beast, but a 9/10 chance of beating Mouse.
You can choose the order of your bouts from these two options:
Mouse, Beast, Mouse
Beast, Mouse, Beast
In order to have the greatest chance of winning entry into the Golden Cage Club, which option should you choose?
121
TYING THE GRASS AND TYING THE KNOT
In the rural Soviet Union, girls used to play the following game. One girl would take six blades of grass and put them in her hand like so.
Another girl would tie the six top ends into three pairs, and then tie the six bottom ends into three pairs. When the fist was unclasped, if it revealed that the grass was joined in one big loop, it was said that the girl would marry within a year.
In this game, is it more likely than not that the girl will be told that she will marry within a year?
The search for love involves risk, luck, and probabilities. The following puzzle emerged as a problem about numbers on paper slips, but it would later became a problem about rings on fingers.
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THE THREE SLIPS OF PAPER
Three numbers are chosen at random and written on three slips of paper. The numbers can be anything, including fractions and negative numbers. The slips are placed face down on a table. The objective of the game is to identify the slip with the highest number.
Your first move is to select a slip and turn it over. Once you see the number, you can nominate this slip as having the highest number. If you do, the game is over. However, you also have the opportunity to discard the first slip and turn over one of the remaining slips. As before, once you see the number on that slip, you can nominate it as being the highest. Or you can discard it, which means that you must choose the final slip, and nominate that as having the highest number.
The odds that the first slip you turn over has the highest number is 1 in 3, because you are randomly picking one option in three. What is the strategy, however, for the remaining slips that improves your odds of guessing the one with the highest number from 1 in 3?
A follow-up question: there are only two slips of paper. Each one has a number on it that was chosen at random. The slips are face down on a table. You are allowed to turn over only one of them, before stating your guess about which slip has the highest number.
Can you improve your odds from 1 in 2 of naming the slip with the highest number?
The answer to both these questions is yes, you can improve your odds with the right strategy. In the case of the problem with two slips only, it’s a stunning result, one of the most remarkable in this book. The values on the two slips can be any numbers at all. You only see one of them. Yet it is possible to guess which of the two is the biggest and have a greater than 50 per cent chance of being correct. Wow.
But what has all this got to do with marriage?
Like the Tuesday-boy problem, the roots of the slips of paper puzzle can be traced back to Martin Gardner and his Scientific American column, which ran monthly from 1959 to 1980. In 1960, he described a variation of the paper slip puzzle that allowed for as many slips as you like. The table now contains a multitude of face-down slips, each with a different random number on it. Again the aim is to name the highest number. You are allowed to turn over any one slip, but as soon as you do you must either choose it (and forgo seeing the others) or discard it. (The solution for that particular version is also in the back.)
Gardner suggested that this game models a woman’s search for a husband. Imagine a woman decides she will date, say, 10 men, each of different suitability, in a year, and imagine that she will indicate after each date whether he should propose. (One assumes that if she gives the okay he will propose, and she will say yes.) The men are the paper slips and the number is their ranking in terms of suitability. Clearly, the woman wants to marry the husband who ranks highest (i.e., she wants to choose the highest number). Once she has started dating, the following rules apply: if she allows someone to propose, she is not allowed to go on a date with the remaining men. And if she rejects a guy, then she cannot return to him later. This behaviour models having to either choose a slip and forgo seeing the others, or reject it and move on. Whether or not one agrees with Gardner’s simplifications – and even he took them with a pinch of salt – the puzzle has become known as the ‘marriage problem’. The field of maths involved here is called ‘optimal stopping’: in a process that could conceivably go on and on, based on what has happened in the past, when is it in your best interests to stop?
The following teaser also first appeared in Martin Gardner’s Scientific American column, in 1959. It was to become the most notorious probability puzzle of the last 50 years.
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THE THREE PRISONERS
Three prisoners, A, B and C, are to be executed next week. In a moment of benevolence, the prison governor picks one at random to be pardoned. The governor visits the prisoners individually to tell them of her decision, but adds that she will not say which prisoner will be pardoned until the day of execution.
Prisoner A, who considers himself the smartest of the bunch, asks the governor the following question: ‘Since you won’t tell me who will be pardoned, could you instead tell me the name of a prisoner who will be executed? If B will be pardoned, give me C’s name. If C will be pardoned, give me B’s name. And if I am to be pardoned, flip a coin to decide whether to tell me B or C.’
The governor thinks about A’s request overnight. She reasons that telling A the name of a prisoner that will be executed is not giving away the name of the prisoner that will be pardoned, so she decides to comply with his demand. The following day she goes into A’s cell and tells him that B will be executed.
A is overjoyed. If B is definitely going to be executed, then either C or A will be pardoned, meaning that A’s chances of being pardoned have just increased from 1 in 3 to 1 in 2.
A tells C what happened. C is also overjoyed, since he too reasons that his chances of being pardoned have risen from 1 in 3 to 1 in 2.
Did both prisoners reason correctly? And if not, what are their odds of being pardoned?
This puzzle may feel familiar. At its core is Bertrand’s box paradox, which we
discussed only a few pages ago.
It is also equivalent to a much more famous puzzle involving two horned, bearded ruminants, a car and a TV host.
The host is Monty Hall, who presented the US game show Let’s Make a Deal between 1963 and 1990. In the final segment of the show a prize was hidden behind one of three doors and the contestant had to choose the correct door to win it. (Let’s Make a Deal never caught on outside the US. A British version hosted by Mike Smith and Julian Clary ran for a single season in 1989.) Here’s the ‘Monty Hall problem’ in full.
You have made the final round on Let’s Make a Deal. In front of you are three doors. Behind one door is an expensive car; behind the other two doors are goats. Your aim is to choose the door with the car behind it.
The host, Monty Hall, says that once you have made your choice he will open one of the other doors to reveal a goat. (Since he knows where the car is, he can always do this. If the door you choose conceals the car, he will choose randomly between which of the other two doors to open).
The game begins and you pick door No. 1.
Monty Hall opens door No. 2, to reveal a goat.